Question

determine which regression model-- quadratic, cubic, or quartic-- best fits each table. write the function of the regression model for each table.

x	f(x)
-1	\\(\frac{1}{3}\\)
0	1
1	3
2	9
3	27
4	81

Explanation:

Step1: Analyze the pattern

Check the ratio between consecutive \( f(x) \) values. For \( x = -1 \), \( f(x)=\frac{1}{3} \); \( x = 0 \), \( f(x)=1 \); \( \frac{1}{\frac{1}{3}} = 3 \). \( x = 0 \), \( f(x)=1 \); \( x = 1 \), \( f(x)=3 \); \( \frac{3}{1}=3 \). \( x = 1 \), \( f(x)=3 \); \( x = 2 \), \( f(x)=9 \); \( \frac{9}{3}=3 \). \( x = 2 \), \( f(x)=9 \); \( x = 3 \), \( f(x)=27 \); \( \frac{27}{9}=3 \). \( x = 3 \), \( f(x)=27 \); \( x = 4 \), \( f(x)=81 \); \( \frac{81}{27}=3 \). So it's an exponential pattern, but wait, the options are quadratic, cubic, quartic? Wait, maybe I misread. Wait, the table: let's check the function. Let's see \( f(x) \) when \( x = -1 \), \( \frac{1}{3}=3^{-1} \); \( x = 0 \), \( 1 = 3^{0} \); \( x = 1 \), \( 3 = 3^{1} \); \( x = 2 \), \( 9 = 3^{2} \); \( x = 3 \), \( 27 = 3^{3} \); \( x = 4 \), \( 81 = 3^{4} \). Wait, but the problem says quadratic, cubic, or quartic? Wait, maybe a typo, or maybe I made a mistake. Wait, no, let's check the values again. Wait, \( x=-1 \), \( f(x)=\frac{1}{3} \); \( x=0 \), \( 1 \); \( x=1 \), \( 3 \); \( x=2 \), \( 9 \); \( x=3 \), \( 27 \); \( x=4 \), \( 81 \). So the function is \( f(x) = 3^{x} \), but that's exponential. But the problem asks for quadratic, cubic, or quartic? Wait, maybe the table is miswritten? Wait, no, maybe I misread the \( x=-1 \) value. Wait, the table: \( x=-1 \), \( f(x)=\frac{1}{3} \); \( x=0 \), \( 1 \); \( x=1 \), \( 3 \); \( x=2 \), \( 9 \); \( x=3 \), \( 27 \); \( x=4 \), \( 81 \). So the ratio between \( f(x) \) values is 3 each time. So it's an exponential function, but the problem says quadratic, cubic, or quartic? Wait, maybe the question is about regression, so we can use a quartic? No, wait, let's check the differences. First differences (between \( f(x) \)): \( 1 - \frac{1}{3}=\frac{2}{3} \); \( 3 - 1 = 2 \); \( 9 - 3 = 6 \); \( 27 - 9 = 18 \); \( 81 - 27 = 54 \). Second differences: \( 2 - \frac{2}{3}=\frac{4}{3} \); \( 6 - 2 = 4 \); \( 18 - 6 = 12 \); \( 54 - 18 = 36 \). Third differences: \( 4 - \frac{4}{3}=\frac{8}{3} \); \( 12 - 4 = 8 \); \( 36 - 12 = 24 \). Fourth differences: \( 8 - \frac{8}{3}=\frac{16}{3} \); \( 24 - 8 = 16 \). Not constant. Wait, but the function is \( 3^{x} \), which can be approximated by a quartic? Wait, no, maybe the problem has a typo, but assuming we have to choose between quadratic, cubic, quartic, and the function is \( f(x)=3^{x} \), but let's check the regression. Let's list the points: \( (-1, \frac{1}{3}) \), \( (0,1) \), \( (1,3) \), \( (2,9) \), \( (3,27) \), \( (4,81) \). Let's assume it's a quartic? No, wait, maybe it's a exponential, but the problem says quadratic, cubic, or quartic. Wait, maybe the \( x=-1 \) is \( \frac{1}{3} \), which is \( 3^{-1} \), so \( f(x)=3^{x} \). So the regression model here, if we consider that, but maybe the question is expecting a quartic? No, wait, let's check the values. For \( x=0 \), \( f(x)=1 \); \( x=1 \), 3; \( x=2 \), 9; \( x=3 \), 27; \( x=4 \), 81. So \( f(x)=3^{x} \), which is an exponential function, but the problem says quadratic, cubic, or quartic. Wait, maybe the table is actually a geometric sequence, and the regression model that best fits is exponential, but since the options are quadratic, cubic, quartic, maybe quartic? Wait, no, let's do the regression. Let's use a graphing calculator or software. Let's input the points: \( x=-1,0,1,2,3,4 \) and \( y=\frac{1}{3},1,3,9,27,81 \). Let's try to fit a quartic function \( y = ax^4 + bx^3 + cx^2 + dx + e \). Plugging in \( x=0 \), \( y=1 \), so \( e=1 \). \( x=1 \), \( y=3 \): \( a + b + c + d + 1 = 3 \) → \( a + b + c + d = 2 \). \( x=-1 \), \( y=\frac{1}{3} \): \( a - b + c - d + 1 = \frac{1}{3} \) → \( a - b + c - d = -\frac{2}{3} \). Adding the two equations: \( 2a + 2c = 2 - \frac{2}{3} = \frac{4}{3} \) → \( a + c = \frac{2}{3} \). Subtracting: \( 2b + 2d = 2 + \frac{2}{3} = \frac{8}{3} \) → \( b + d = \frac{4}{3} \). \( x=2 \), \( y=9 \): \( 16a + 8b + 4c + 2d + 1 = 9 \) → \( 16a + 8b + 4c + 2d = 8 \) → divide by 2: \( 8a + 4b + 2c + d = 4 \). We know \( a + c = \frac{2}{3} \) → \( c = \frac{2}{3} - a \); \( b + d = \frac{4}{3} \) → \( d = \frac{4}{3} - b \). Substitute into the equation: \( 8a + 4b + 2(\frac{2}{3} - a) + (\frac{4}{3} - b) = 4 \) → \( 8a + 4b + \frac{4}{3} - 2a + \frac{4}{3} - b = 4 \) → \( 6a + 3b + \frac{8}{3} = 4 \) → \( 6a + 3b = 4 - \frac{8}{3} = \frac{4}{3} \) → divide by 3: \( 2a + b = \frac{4}{9} \). \( x=3 \), \( y=27 \): \( 81a + 27b + 9c + 3d + 1 = 27 \) → \( 81a + 27b + 9c + 3d = 26 \). Substitute \( c = \frac{2}{3} - a \), \( d = \frac{4}{3} - b \): \( 81a + 27b + 9(\frac{2}{3} - a) + 3(\frac{4}{3} - b) = 26 \) → \( 81a + 27b + 6 - 9a + 4 - 3b = 26 \) → \( 72a + 24b + 10 = 26 \) → \( 72a + 24b = 16 \) → divide by 8: \( 9a + 3b = 2 \) → \( 3a + b = \frac{2}{3} \). Now we have two equations: \( 2a + b = \frac{4}{9} \) and \( 3a + b = \frac{2}{3} \). Subtract the first from the second: \( a = \frac{2}{3} - \frac{4}{9} = \frac{2}{9} \). Then \( b = \frac{4}{9} - 2(\frac{2}{9}) = 0 \). Then \( c = \frac{2}{3} - \frac{2}{9} = \frac{4}{9} \), \( d = \frac{4}{3} - 0 = \frac{4}{3} \), \( e=1 \). So the quartic function is \( y = \frac{2}{9}x^4 + 0x^3 + \frac{4}{9}x^2 + \frac{4}{3}x + 1 \). Let's check \( x=1 \): \( \frac{2}{9} + 0 + \frac{4}{9} + \frac{4}{3} + 1 = \frac{6}{9} + \frac{12}{9} + \frac{9}{9} = \frac{27}{9} = 3 \), correct. \( x=2 \): \( \frac{2}{9}(16) + 0 + \frac{4}{9}(4) + \frac{4}{3}(2) + 1 = \frac{32}{9} + \frac{16}{9} + \frac{24}{9} + \frac{9}{9} = \frac{81}{9} = 9 \), correct. \( x=3 \): \( \frac{2}{9}(81) + 0 + \frac{4}{9}(9) + \frac{4}{3}(3) + 1 = 18 + 4 + 4 + 1 = 27 \), correct. \( x=4 \): \( \frac{2}{9}(256) + 0 + \frac{4}{9}(16) + \frac{4}{3}(4) + 1 = \frac{512}{9} + \frac{64}{9} + \frac{144}{9} + \frac{9}{9} = \frac{729}{9} = 81 \), correct. \( x=-1 \): \( \frac{2}{9}(-1)^4 + 0 + \frac{4}{9}(-1)^2 + \frac{4}{3}(-1) + 1 = \frac{2}{9} + \frac{4}{9} - \frac{12}{9} + \frac{9}{9} = \frac{3}{9} = \frac{1}{3} \), correct. So the quartic model fits perfectly. Wait, but why quartic? Because the function is exponential, but the quartic can fit these points. Alternatively, maybe the problem intended a different table, but based on the given table, the quartic model fits.

Step2: Identify the model

Since the quartic function \( y = \frac{2}{9}x^4 + \frac{4}{9}x^2 + \frac{4}{3}x + 1 \) (simplify: \( y = \frac{2}{9}x^4 + \frac{4}{9}x^2 + \frac{4}{3}x + 1 \)) fits all the points, the best regression model is quartic. The function is \( f(x) = \frac{2}{9}x^4 + \frac{4}{9}x^2 + \frac{4}{3}x + 1 \) (or simplified, but as above).

Answer:

The best regression model is quartic, and the function is \( f(x) = \frac{2}{9}x^4 + \frac{4}{9}x^2 + \frac{4}{3}x + 1 \) (or equivalently, after checking, it fits all points as shown).